Question

If the self inductance of 500 turns coil is 125mH, then the self inductance of the similar coil of $800$ turns
$\begin{align}
  & A.48.8mH \\
 & B.200mH \\
 & C.290mH \\
 & D.320mH \\
\end{align}$

Answer 1

Hint: Here we need to see what are the factors on which the self inductance of a coil depends.We are given two numbers of turns of the coil and the value of self inductance corresponding to one number of turns. As all the parameters are remaining the same so knowing the dependence of self inductance only on the number of turns will be enough to solve the question.

Formula used:
$L=\dfrac{{{\mu }_{0}}\pi {{N}^{2}}r}{2},\dfrac{{{L}_{1}}}{{{L}_{2}}}=\dfrac{N_{1}^{2}}{N_{2}^{2}}$

Complete step by step solution:
The self inductance $L$ of a coil of radius $r$ ,number of turns $N$ is given by
$L\propto {{N}^{2}}$ $L=\dfrac{{{\mu }_{0}}\pi {{N}^{2}}r}{2}$, where ${{\mu }_{0}}$ is the permeability of free space. Now if all the other parameters except $N$ remain constant, then we can write
$L\propto {{N}^{2}}$ . Now in this problem ${{L}_{1}}=125mH$ ,${{N}_{1}}=500$ ,${{N}_{2}}=800,{{L}_{2}}=?$
Therefore we can write
$\begin{align}
  & \dfrac{{{L}_{1}}}{{{L}_{2}}}=\dfrac{N_{1}^{2}}{N_{2}^{2}} \\
 & or\dfrac{125}{{{L}_{2}}}=\dfrac{{{500}^{2}}}{{{800}^{2}}} \\
 & or{{L}_{2}}=125\times \dfrac{64}{25}=320mH \\
\end{align}$
Thus the self inductance of similar coil of $800$ turns will be $320mH$ .

So the correct option is D.

Additional information:
Self-induction is the phenomenon of production of opposing induced emf in a coil as a result of varying current in the coil itself. The self inductance of a coil is equal to the emf induced in the coil when the rate of change of current in the coil is unity.

Note:
The self inductance of a coil depends on many factor as we can see from the expression
$L=\dfrac{{{\mu }_{0}}\pi {{N}^{2}}r}{2},$ Now in this problem the only quantity that has changed is $N$ , so we are able to apply the relation $L\propto {{N}^{2}}$. If in any other problem, all the quantities are changing then we need to consider all the quantities and accordingly we need to go through the solution.As in this problem the unit of self inductance is given in $mH$ , that is why the answer will also come in $mH.$